Solve (8x - 6)/(3x - 6) = (5x - 6x)/(2 - 6) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{8x-6}{3x-6}=\frac{5x-6x}{2-6}$$

Solve for x

$x=12$
$x=\frac{2}{3}\approx 0.666666667$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $3\left(x-2\right)$.

$$8x-6=\left(-\frac{3}{4}x+\frac{3}{2}\right)\left(5x-6x\right)$$

Combine $5x$ and $-6x$ to get $-x$.

$$8x-6=\left(-\frac{3}{4}x+\frac{3}{2}\right)\left(-1\right)x$$

Use the distributive property to multiply $-\frac{3}{4}x+\frac{3}{2}$ by $-1$.

$$8x-6=\left(\frac{3}{4}x-\frac{3}{2}\right)x$$

Use the distributive property to multiply $\frac{3}{4}x-\frac{3}{2}$ by $x$.

$$8x-6=\frac{3}{4}x^{2}-\frac{3}{2}x$$

Subtract $\frac{3}{4}x^{2}$ from both sides.

$$8x-6-\frac{3}{4}x^{2}=-\frac{3}{2}x$$

Add $\frac{3}{2}x$ to both sides.

$$8x-6-\frac{3}{4}x^{2}+\frac{3}{2}x=0$$

Combine $8x$ and $\frac{3}{2}x$ to get $\frac{19}{2}x$.

$$\frac{19}{2}x-6-\frac{3}{4}x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-\frac{3}{4}x^{2}+\frac{19}{2}x-6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-\frac{3}{4}$ for $a$, $\frac{19}{2}$ for $b$, and $-6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\frac{19}{2}±\sqrt{\left(\frac{19}{2}\right)^{2}-4\left(-\frac{3}{4}\right)\left(-6\right)}}{2\left(-\frac{3}{4}\right)}$$

Square $\frac{19}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{19}{2}±\sqrt{\frac{361}{4}-4\left(-\frac{3}{4}\right)\left(-6\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply $-4$ times $-\frac{3}{4}$.

$$x=\frac{-\frac{19}{2}±\sqrt{\frac{361}{4}+3\left(-6\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply $3$ times $-6$.

$$x=\frac{-\frac{19}{2}±\sqrt{\frac{361}{4}-18}}{2\left(-\frac{3}{4}\right)}$$

Add $\frac{361}{4}$ to $-18$.

$$x=\frac{-\frac{19}{2}±\sqrt{\frac{289}{4}}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of $\frac{289}{4}$.

$$x=\frac{-\frac{19}{2}±\frac{17}{2}}{2\left(-\frac{3}{4}\right)}$$

Multiply $2$ times $-\frac{3}{4}$.

$$x=\frac{-\frac{19}{2}±\frac{17}{2}}{-\frac{3}{2}}$$

Now solve the equation $x=\frac{-\frac{19}{2}±\frac{17}{2}}{-\frac{3}{2}}$ when $±$ is plus. Add $-\frac{19}{2}$ to $\frac{17}{2}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{1}{-\frac{3}{2}}$$

Divide $-1$ by $-\frac{3}{2}$ by multiplying $-1$ by the reciprocal of $-\frac{3}{2}$.

$$x=\frac{2}{3}$$

Now solve the equation $x=\frac{-\frac{19}{2}±\frac{17}{2}}{-\frac{3}{2}}$ when $±$ is minus. Subtract $\frac{17}{2}$ from $-\frac{19}{2}$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{18}{-\frac{3}{2}}$$

Divide $-18$ by $-\frac{3}{2}$ by multiplying $-18$ by the reciprocal of $-\frac{3}{2}$.

$$x=12$$

The equation is now solved.

$$x=\frac{2}{3}$$ $$x=12$$

Steps for Completing the Square

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $3\left(x-2\right)$.

$$8x-6=\left(-\frac{3}{4}x+\frac{3}{2}\right)\left(5x-6x\right)$$

Combine $5x$ and $-6x$ to get $-x$.

$$8x-6=\left(-\frac{3}{4}x+\frac{3}{2}\right)\left(-1\right)x$$

Use the distributive property to multiply $-\frac{3}{4}x+\frac{3}{2}$ by $-1$.

$$8x-6=\left(\frac{3}{4}x-\frac{3}{2}\right)x$$

Use the distributive property to multiply $\frac{3}{4}x-\frac{3}{2}$ by $x$.

$$8x-6=\frac{3}{4}x^{2}-\frac{3}{2}x$$

Subtract $\frac{3}{4}x^{2}$ from both sides.

$$8x-6-\frac{3}{4}x^{2}=-\frac{3}{2}x$$

Add $\frac{3}{2}x$ to both sides.

$$8x-6-\frac{3}{4}x^{2}+\frac{3}{2}x=0$$

Combine $8x$ and $\frac{3}{2}x$ to get $\frac{19}{2}x$.

$$\frac{19}{2}x-6-\frac{3}{4}x^{2}=0$$

Add $6$ to both sides. Anything plus zero gives itself.

$$\frac{19}{2}x-\frac{3}{4}x^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-\frac{3}{4}x^{2}+\frac{19}{2}x=6$$

Divide both sides of the equation by $-\frac{3}{4}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}x^{2}+\frac{19}{2}x}{-\frac{3}{4}}=\frac{6}{-\frac{3}{4}}$$

Dividing by $-\frac{3}{4}$ undoes the multiplication by $-\frac{3}{4}$.

$$x^{2}+\frac{\frac{19}{2}}{-\frac{3}{4}}x=\frac{6}{-\frac{3}{4}}$$

Divide $\frac{19}{2}$ by $-\frac{3}{4}$ by multiplying $\frac{19}{2}$ by the reciprocal of $-\frac{3}{4}$.

$$x^{2}-\frac{38}{3}x=\frac{6}{-\frac{3}{4}}$$

Divide $6$ by $-\frac{3}{4}$ by multiplying $6$ by the reciprocal of $-\frac{3}{4}$.

$$x^{2}-\frac{38}{3}x=-8$$

Divide $-\frac{38}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{19}{3}$. Then add the square of $-\frac{19}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{38}{3}x+\left(-\frac{19}{3}\right)^{2}=-8+\left(-\frac{19}{3}\right)^{2}$$

Square $-\frac{19}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{38}{3}x+\frac{361}{9}=-8+\frac{361}{9}$$

Add $-8$ to $\frac{361}{9}$.

$$x^{2}-\frac{38}{3}x+\frac{361}{9}=\frac{289}{9}$$

Factor $x^{2}-\frac{38}{3}x+\frac{361}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{19}{3}\right)^{2}=\frac{289}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{3}\right)^{2}}=\sqrt{\frac{289}{9}}$$

Simplify.

$$x-\frac{19}{3}=\frac{17}{3}$$ $$x-\frac{19}{3}=-\frac{17}{3}$$

Add $\frac{19}{3}$ to both sides of the equation.

$$x=12$$ $$x=\frac{2}{3}$$